A uniform electric field of \(2\times10^{3}\ \mathrm{V\,m^{-1}}\) points along the +x direction. A charge of \(+5\ \mu\mathrm{C}\) moves from the point x = 0 to the point x = 30 cm. The change in the potential energy of the charge is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(-3\times10^{-3}\ \mathrm{J}\)
The potential energy of a charge \( q \) at a point in an external field is \( qV \), so its change is \( q\,\Delta V \). The potential falls along the direction of the field. In a uniform field it falls by \( Ed \) over a distance \( d \) measured along the field.
\( \begin{aligned} \Delta V &= -Ed \\ &= -2\times10^{3}\times0.3 \\ &= -600\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} \Delta U &= q\,\Delta V \\ &= 5\times10^{-6}\times(-600) \\ &= -3\times10^{-3}\ \mathrm{J} \end{aligned} \)
The negative sign shows that the potential energy decreases as the positive charge moves along the field.
NCERT (2026-27):
Potential energy of a single charge – “Potential energy of q at r in an external field = qV(r)”
\( U=qV(\mathbf{r}) \qquad (2.27) \)
Relation between field and potential – “(i) Electric field is in the direction in which the potential decreases steepest.”